GCC

题目描述

The GNU Compiler Collection (usually shortened to GCC) is a compiler system produced by the GNU Project supporting various programming languages.  But it doesn’t contains the math operator “!”.

In mathematics the symbol represents the factorial operation. The expression n! means "the product of the integers from 1 to n". For example, 4! (read four factorial) is 4 × 3 × 2 × 1 = 24. (0! is defined as 1, which is a neutral element in multiplication, not multiplied by anything.)

We want you to help us with this formation: (0! + 1! + 2! + 3! + 4! + ... + n!)%m. 

输入

The first line consists of an integer T, indicating the number of test cases.

Each test on a single consists of two integer n and m. 

0 < T <= 20
0 <= n < 10 ¹⁰⁰ (without leading zero) 
0 < m < 1000000 

输出

Output the answer of (0! + 1! + 2! + 3! + 4! + ... + n!)%m. 

示例输入

1 

10 861017 

示例输出

593846

题意：输入n和m，计算(0! + 1! + 2! + 3! + 4! + ... + n!)%m的值。。。
思路：因为看到 n 最大是10^100，顿时感觉很蒙，可是，只要仔细想想，n这么大是唬人的。因为只要n>=m时，它的阶乘对m取余都是0，也就相当于 (0! + 1! + 2! + 3! + 4! + ... + （m-1)!)%m.
　　　若原本n就小于m，直接算就行了。就是分步取余。

 1 #include<stdio.h>

 2 #include<string.h>

 3 

 4 int MOD(int n,int m)

 5 {

 6     long long ans = 1,tmp = 1;

 7     int num;

 8     for(num = 1; num <= n; num++)

 9     {

10         tmp = (tmp*(num%m))%m;

11         ans = (ans+tmp)%m;

12     }

13     ans = ans%m;

14     return ans;

15 }

16 

17 int main()

18 {

19     int test;

20     char s[110];

21     int n,m,ans;

22     scanf("%d",&test);

23     while(test--)

24     {

25         scanf("%s %d",s,&m);

26         if(m == 1)

27         {

28             printf("0\n");

29             continue;

30         }

31         n = 0;

32         for(int i = 0; s[i]; i++)

33         {

34             n = n*10+s[i]-'0';

35             if(n >= m)

36             {

37                 n = m-1;

38                 break;

39             }

40         }

41         ans = MOD(n,m);

42         printf("%d\n",ans);

43     }

44     return 0;

45 }

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